CHAPTER 5 Work and Energy http://www.physicsclassroom.com/Class/energy/ energtoc.html Units

• • • • • • • • • •

Work Done by a Constant Force Work Done by a Varying Force Kinetic Energy, and the Work-Energy Principle Potential Energy Conservative and Nonconservative Forces Mechanical Energy and Its Conservation Problem Solving Using Conservation of Mechanical Energy Other Forms of Energy; Energy Transformations and the Law of Conservation of Energy Energy Conservation with Dissipative Forces: Solving Problems Power Work Done by a Constant Force

The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement: •

• • • •

This gives no information about – the time it took for the displacement to occur – the velocity or acceleration of the object Work is a scalar quantity The work done by a force is zero when the force is perpendicular to the displacement – cos 90° = 0 If there are multiple forces acting on an object, the total work done is the algebraic sum of the amount of work done by each force Work can be positive or negative – Positive if the force and the displacement are in the same direction – Negative if the force and the displacement are in the opposite direction In the SI system, the units of work are joules: As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion. 1

The work done may be positive, zero, or negative, depending on the angle between the force and the displacement:

• •

Work is positive when lifting the box Work would be negative if lowering the box – The force would still be upward, but the displacement would be downward

Work done by forces that oppose the direction of motion, such as friction, will be negative. Centripetal forces do no work, as they are always perpendicular to the direction of motion.

For a force that varies, the work can be approximated by dividing the distance up into small pieces, finding the work done during each, and adding them up. As the pieces become very narrow, the work done is the area under the force vs. distance curve.

Example 1: A person pulls a 50-kg crate 40 m along a horizontal floor by a constant force FP  110 N , which acts at a 37o angle. The floor

2

is rough and exerts a friction force Ffr  50 N . Determine (a) the work done by each force acting on the crate.

WP  FP x cos   (110 N )(40m) cos37o  3514 J W fr  Ffr x cos180o  (50 N )(40m)(1)  2000 J

Determine (b) the net work done on the crate.

Wnet  Wmg  WN  WP  W fr Wnet  0  0  3514 J  (2000 J )  1514 J Example 2: An intern pushes a 72-kg patient on a 15-kg gurney, producing an acceleration of 0.60m / s 2 . How much work does the intern do by pushing the patient and gurney through a distance of 2.5m? Assume the gurney moves without friction. F  ma  (72kg  15kg )(0.60m / s 2 )  52N W  Fd  (52 N )(2.5m)  130 J Kinetic Energy and the Work-Energy Principle Energy was traditionally defined as the ability to do work. We now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition. If we write the acceleration in terms of the velocity and the distance, we find that the work done here is We define the kinetic energy:

KE 

1 2 mv 2

This means that the work done is equal to the change in the kinetic energy:

Wnet  KE • •

If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases.

3

Because work and kinetic energy can be equated, they must have the same units: kinetic energy is measured in joules.

Example 3: A 145-g baseball is thrown so that it acquires a speed of 25 m/s. (a) What is its kinetic energy? 1 1 KE  mv 2  (0.145kg )(25m / s) 2  45 J 2 2 (b) What is the net work done on the ball to make it reach this speed, if it started from rest? Since the initial KE was zero, the net work done is just equal to the final KE, 45J. Example 4: How much net work is required to accelerate a 1000-kg car from 20 m/s to 30 m/s?

1 1 W  KE2  KE1  mv22  mv12 2 2 1 1 2  (1000kg )(30m / s)  (1000kg )(20m / s)2  2.5 x105 J 2 2

Example 5: The driver of a 1000-kg car traveling at 35.0 m/s slams on his brakes to avoid hitting another vehicle in front of him. After the brakes are applied, a constant friction force of 8000N acts on the car. (a) At what minimum distance should the brakes be applied to avoid a collision? Apply the work energy theorem:

1 1 Wnet  mv 2f  mvi2 2 2

Substitute an expression for the frictional work:

1  Fkf x  0  mvi2 2

1 (8000 N )x  0  (1000kg )(35.0m / s) 2 2

x  76.6m

4

(b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collision occur? 1 1 Wnet  W fr   Fkf x  mv 2f  mvi2 2 2

v 2f  vi2 

2 Fkf x m

 2  2 2 v2f  (35.0m / s)2    (8000 N )(30.0m)  745m / s v f  27.3m / s 1000 kg   Potential Energy An object can have potential energy by virtue of its surroundings. Familiar examples of potential energy: • A wound-up spring • A stretched elastic band • An object at some height above the ground In raising a mass m to a height h, the work done by the external force is

We therefore define the gravitational potential energy:

PEgrav  mgy •

• •

Gravitational Potential Energy is the energy associated with the relative position of an object in space near the Earth’s surface – Objects interact with the earth through the gravitational force – Actually the potential energy is for the earth-object system PE = mgy Wgravity  PEi  PE f

Units of Potential Energy are the same as those of Work and Kinetic Energy This potential energy can become kinetic energy if the object is dropped. Potential energy is a property of a system as a whole, not just of the object (because it depends on external forces). If PEgrav  mgy , where do we measure y from? •

5

It turns out not to matter, as long as we are consistent about where we choose y = 0. Only changes in potential energy can be measured. Problem Solving Using Conservation of Mechanical Energy

In the image on the left, the total mechanical energy is:

The energy buckets (right) show how the energy moves from all potential to all kinetic.

If there is no friction, the speed of a roller coaster will depend only on its height compared to its starting height.

Example 6: A 60.0kg skier is at the top of a slope. At the initial point (A), she is 10.0m vertically above point (B). Setting the zero level for gravitational potential energy at (B), find the gravitational potential energy of this system when the skier is at (A).

PEi  mgyi  (60.0kg )(9.80m / s 2 )(10.0m)  5.88x103 J

Example 7: A diver of mass m drops from a board 10.0 m above the water’s surface. Neglect air resistance. (a) Use conservation of mechanical energy to find his speed 5.00 m above the water’s surface.

6

KEi  PEi  KE f  PE f

1 2 1 mvi  mgyi  mv 2f  mgy f 2 2 v f  2 g ( yi  y f )

1 0  gyi  v 2f  gy f 2

 2(9.80m / s 2 )(10.0m  5.0m  9.90m / s

(b) Find his speed as he hits the water.

1 0  mgyi  mv 2f  0 2

v f  2 gyi  2(9.80m / s ) 2 (10.0m)  14.0m / s

Example 8: A grasshopper launches itself at an angle of 45o above the horizontal and rises to a maximum height of 1.00 m. With what speed did it leave the ground?

vi  2 gh  2 (9.80m / s 2 )(1.00m)  6.26m / s Potential Energy Potential energy can also be stored in a spring when it is compressed; the figure below shows potential energy yielding kinetic energy.

• •

Involves the spring constant, k Hooke’s Law gives the force – F=-kx • F is the restoring force • F is in the opposite direction of x • k depends on how the spring was formed, the material it is made from, thickness of the wire, etc. The force required to compress or stretch a spring is:

where k is called the spring constant, and needs to be measured for each spring.

7

The force increases as the spring is stretched or compressed further. We find that the potential energy of the compressed or stretched spring, measured from its equilibrium position, can be written:

1 elastic PE  kx 2 2

Conservative and Nonconservative Forces If friction is present, the work done depends not only on the starting and ending points, but also on the path taken. Friction is called a nonconservative force.

Potential energy can only be defined for conservative forces.

Problem Solving Using Conservation of Mechanical Energy For an elastic force, conservation of energy tells us:

8

Example 9: A dart of mass 0.100 kg is pressed against the spring of a toy dart gun. The spring (k = 250 N/m) is compressed 6.0cm and released. If the dart detaches from the spring when the spring reaches its natural length (x = 0) what speed does the dart acquire?

1 1 0  kx12  mv22  0 2 2 v22 

kx12 (250 N / m)(0.060m) 2   9.0m2 / s 2 m (0.100kg )

v  3.0m / s

Example 10: A ball of mass = 2.60 kg, starting from rest, falls a vertical distance h = 55.0 cm before striking a vertical coiled spring, which compresses an amount Y = 15.0 cm. Determine the spring constant.

k

2mg (h  Y ) 2(2.60kg )(9.80m / s 2 )(0.550m  0.150m)  Y2 (0.150m) 2

 1590 N / m

Example 11: A block with mass of 5.00 kg is attached to a horizontal spring with spring constant k = 400 N/m. The surface the block rests upon is frictionless. If the block is pulled out to xi = 0.50 m and released, (a) Find the speed of the block at the equilibrium point.

vf 

k xi m



400 N / m (0.05m)  0.447m / s 5.0kg

(b) Find the speed when x = 0.025 m.

400 N / m k 2  [(0.050m)2  (0.025m)2 ]  0.387m / s ( xi  x 2f ) 5.0kg m (c) If there is friction in part (a) what is the speed? k 2 400 N / m vf xi  2k gxi  (0.05m)2  2(0.150)(9.80m / s 2 )(0.05m)  0.23m / s m 5.0kg vf 

9

Example 12: A 0.50kg block rests on a horizontal, frictionless surface. The block is pressed back against a spring having a constant of k = 625N/m, compressing the spring by 10 cm to point (A). Then the block is released. (a) Find the maximum distance d the block travels up the frictionless incline if   30o .

1 2 1 kxi (625 N / m)(0.10m) 2 2 d 2   1.28m mg sin  (0.50kg )(9.80m / s 2 )sin 30.0o (b) How fast is the block going when halfway to its maximum height?

vf 

k 2 xi  gh m

 625N / m  2 2    (0.10m)  (9.80m / s )(0.640m)  2.50m / s 0.50 kg  

Other Forms of Energy; Energy Transformations and the Conservation of Energy Some other forms of energy: Electric energy, nuclear energy, thermal energy, chemical energy. Work is done when energy is transferred from one object to another. Accounting for all forms of energy, we find that the total energy neither increases nor decreases. Energy as a whole is conserved. Energy Conservation with Dissipative Processes; Solving Problems If there is a nonconservative force such as friction, where do the kinetic and potential energies go? They become heat; the actual temperature rise of the materials involved can be calculated. Power is the rate at which work is done –

Power

In the SI system, the units of power are watts:

The difference between walking and running up these stairs is power – the change in gravitational potential energy is the same.

10

Power is also needed for acceleration and for moving against the force of gravity. The average power can be written in terms of the force and the average velocity:

•

US Customary units are generally hp – Need a conversion factor

ft lb  746 W s – Can define units of work or energy in terms of units of power: • kilowatt hours (kWh) are often used in electric bills • This is a unit of energy, not power 1 hp  550

Example 13: A 60-kg jogger runs up a long flight of stairs in 4.0s. The vertical height of the stairs is 4.5 m. (a) Estimate the jogger’s power output in watts and horsepower.

P



W mgy  t t

(60kg )(9.80m / s 2 )(4.5m)  660W 4.0s 660W  0.88hp 746W / hp

(b) How much energy did this require? E  Pt

P  660W  660 J / s

E  (660 J / s)(4.0s)  2600 J

Example 14: Calculate the power required of a 1400-kg car under the following circumstances: (a) The car climbs a 10 o hill at a steady 80 km/h;

F  700 N  mg sin10o  700N  (1400kg )(9.80m / s 2 )(0.174)  3100N P  Fv  (3100 N )(22m / s)  6.80 x104W  91hp

11

(b) The car accelerates along a level road from 90 to 110 km/h in 6.0 s to pass another car. Assume the retarding force on the car is FR  700N throughout.

a

v  vo (30.6m / s  25.0m / s)   0.93m / s 2 t 6.0s

F  max  FR  (1400kg )(0.93m / s 2 )  700 N

 1300 N  700 N  2000 N

P  Fv  (2000 N )(30.6m / s)  6.12 x104W  82hp Example 15: A 1000 kg elevator carries a maximum load of 800 kg. A constant frictional force of 4000 N retards its motion upward. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s?  F  ma

T  F fr  Mg  0 T  F fr  Mg

T  4000N  (1800kg )(9.80m / s 2 ) T  2.16 x104 N P  Fv  (2.16x104 N )(3.00m / s)  6.48x104W

P  64.8kW  86.9hp

Example 16: What average power would a 1000 kg speedboat need to go from rest to 20.0 m/s in 5.00 s, assuming the water exerts a constant drag force of Fd  500N and the acceleration is constant?

v  vo  at

(20.0m / s)  0  a(5.00s)  a  4.00m / s 2

v 2  vo2  2ax

(20.0m / s)2  02  2(4.00m / s 2 )x  x  50.0m

Wd   Fd x  (500 N )(50.0m)  2.50 x104 J

P

Wengine t



Wengine  2.25 x105 J

2.25x105 J  4.50 x104W  60.3hp 5.00s

12

CHAPTER 6 WORK AND ENERGY CONCEPTS 1. Two men, Tom and Jerry, push against a wall. Jerry stops after 10 minutes, while Joel is able to push for 5 minutes longer. Based upon this neither do any work. 2. The area under the curve, on a Force-position (F – x) graph, represents work. 3. The rate of change of work with respect to time is called power. 4. A student running up a flight of stairs increases her speed at a constant rate. The graph that best represents the relationship between work and time for the student’s run up the stairs is B.

5. Pulling a 9.8-N cart a distance of 0.50 m along a plane inclined at 15o requires 1.3 J of work. If the cart were raised 0.50 m vertically instead of being pulled along the inclined plane, the amount of work done would be greater. 6. The rate of change of work with respect to time is called power. 7. As the time required to do a given quantity of work decreases, the power developed increases. 8. A rock is dropped from a cliff. Comparing its kinetic energy KE, to its potential energy PE, KE increases and PE decreases.

9. The graph that best represents the relationship between PE and height above ground (h) for a freely falling object from rest is B.

13

10. As the speed of a bicycle moving along a level horizontal surface changes from 2 m/s to 4 m/s, the magnitude of the bicycle’s gravitational PE remains the same.

11. If the velocity of an automobile is doubled, its KE quadruples.

KE 

1 2 mv 2

12. The cart on the right with the greatest KE is A.

13. The graph on the right represents the velocity – time Relationship for a 2.0-kg mass moving along a horizontal frictionless surface. The KE of the mass is greatest during interval A.

14. The graph that best represents the relationship between the PE stored in a spring and the change in the length of the spring from its equilibrium function of its velocity (v) is B.

15. The graph that best represents the relationship between the elongation of an ideal spring and the applied force is A.

14

16. The graph that best represents the relationship between the PE stored in a spring and the change in the length of the spring from its equilibrium position (X) is B.

17. A force is applied to a block, causing it to accelerate along a horizontal, frictionless surface. The energy gained by the block is equal to the work done on the block.

18. The work done in raising an object must result in an increase in the objects gravitational potential energy. 19. As an object falls freely near the Earth’s surface, the loss in gravitational potential energy of the object is equal to its gain in kinetic energy. 20. At the halfway point between the start and the end of an objects fall, the kinetic energy of a freely falling object equals its potential energy. 21. A container of water is lifted vertically 3 m, and then returned to its original position. If the total weight is 30 N the amount of work done was zero. 22. You and your friend want to go to the top of the Eiffel Tower. Your friend takes the elevator straight up. You are macho and walk up the spiral stairway, taking longer to do so. Compared to the gravitational potential energy of you and your friend after reaching the top, you both have the same amount of potential energy.

23. King Kong falls from the top of the Empire State Building, through the air (air friction is present), to the ground below. His kinetic energy (KE) just before striking the ground compared to his potential energy (PE) at the top of the building is KE